Electrostatics of Conductors (NCERT) - NEET NCERT Based Tests Free MCQ

If a conductor has a non zero potential and there are no charge anywhere else outside, then there must be charge on the surface of the conductor or inside the conductor. There cannot be any charge in the body of the conductor.

Since the electric field out of a conductor is perpendicular to the surface. Therefore, it will not have component along the surface of the conductor.
So, there will be no work done in moving the conductor along the surface.
So, it behaves as an equipotential surface.
The charge carried by the conductor is also uniformly distributed over the surface of the conductor.

Here V₁ = V₂ or 
∴ q₁q₂ = R₁R₂ ......(i)
Given R₁ > R₂
∴ q₁ > q₂
∴ Larger sphere has more charge than the smaller sphere. Now charge densities.


As R₁ > R₂ therefore σ₂ > σ₁
Charge density of smaller sphere is more thatn the charge density of large sphere.

Before contact, Charges of each sphered,
q₁​ = σ4πR²; q₂ ​= σ4π(2R)² = 4q_1​
When the two sphere are brought in contact, their charges are shared till their potentials become equal i.e., V_1 ​= V₂ ​

∴ q₂′ ​= 2q₁′​....(i)
As there is no loss of charge in the process
∴ q₁′ ​+ q₂′ ​= q₁ ​+ q₂ ​= q₁ ​+ 4q₁ ​= 5q₁ ​ = 5(σ4πR²)
or q₁′ ​+ 2q₁′ ​= 5σ4πR² (using (i))

Let q₁ and q₂ be the charges and C₁ and C₂ be the capacitance of two spheres
The charge flows from the sphere at higher potential to the other at lower potential, till their potentials becomes equal.
After sharing, the charges on two spheres would be
q₁/q₂ = C₁V/C₂V…(i)
Also C₁C₂ = ab…(ii)
From (i) q₁/q₂ = a/b
Ratio of surface charge on the two spheres
σ₁/σ₂ = q₁/4πa²⋅4πb²/q₂ = q₁/q₂⋅b₂/a₂ = b/a(using(ii)
∴ The ratio of electric fields at the surfaces of two spheres E₁/E₂ = σ₁/σ₂ = b/a